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How to calculate number of days between two dates?
(42 answers)
Closed 3 months ago.
I want to calculate date difference in days, hours, minutes, seconds, milliseconds, nanoseconds. How can I do it?
Assuming you have two Date objects, you can just subtract them to get the difference in milliseconds:
var difference = date2 - date1;
From there, you can use simple arithmetic to derive the other values.
var DateDiff = {
inDays: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return Math.floor((t2-t1)/(24*3600*1000));
},
inWeeks: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000*7));
},
inMonths: function(d1, d2) {
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
return (d2M+12*d2Y)-(d1M+12*d1Y);
},
inYears: function(d1, d2) {
return d2.getFullYear()-d1.getFullYear();
}
}
var dString = "May, 20, 1984";
var d1 = new Date(dString);
var d2 = new Date();
document.write("<br />Number of <b>days</b> since "+dString+": "+DateDiff.inDays(d1, d2));
document.write("<br />Number of <b>weeks</b> since "+dString+": "+DateDiff.inWeeks(d1, d2));
document.write("<br />Number of <b>months</b> since "+dString+": "+DateDiff.inMonths(d1, d2));
document.write("<br />Number of <b>years</b> since "+dString+": "+DateDiff.inYears(d1, d2));
Code sample taken from here.
Another solution is convert difference to a new Date object and get that date's year(diff from 1970), month, day etc.
var date1 = new Date(2010, 6, 17);
var date2 = new Date(2013, 12, 18);
var diff = new Date(date2.getTime() - date1.getTime());
// diff is: Thu Jul 05 1973 04:00:00 GMT+0300 (EEST)
console.log(diff.getUTCFullYear() - 1970); // Gives difference as year
// 3
console.log(diff.getUTCMonth()); // Gives month count of difference
// 6
console.log(diff.getUTCDate() - 1); // Gives day count of difference
// 4
So difference is like "3 years and 6 months and 4 days". If you want to take difference in a human readable style, that can help you.
Expressions like "difference in days" are never as simple as they seem. If you have the following dates:
d1: 2011-10-15 23:59:00
d1: 2011-10-16 00:01:00
the difference in time is 2 minutes, should the "difference in days" be 1 or 0? Similar issues arise for any expression of the difference in months, years or whatever since years, months and days are of different lengths and different times (e.g. the day that daylight saving starts is 1 hour shorter than usual and two hours shorter than the day that it ends).
Here is a function for a difference in days that ignores the time, i.e. for the above dates it returns 1.
/*
Get the number of days between two dates - not inclusive.
"between" does not include the start date, so days
between Thursday and Friday is one, Thursday to Saturday
is two, and so on. Between Friday and the following Friday is 7.
e.g. getDaysBetweenDates( 22-Jul-2011, 29-jul-2011) => 7.
If want inclusive dates (e.g. leave from 1/1/2011 to 30/1/2011),
use date prior to start date (i.e. 31/12/2010 to 30/1/2011).
Only calculates whole days.
Assumes d0 <= d1
*/
function getDaysBetweenDates(d0, d1) {
var msPerDay = 8.64e7;
// Copy dates so don't mess them up
var x0 = new Date(d0);
var x1 = new Date(d1);
// Set to noon - avoid DST errors
x0.setHours(12,0,0);
x1.setHours(12,0,0);
// Round to remove daylight saving errors
return Math.round( (x1 - x0) / msPerDay );
}
This can be more concise:
/* Return number of days between d0 and d1.
** Returns positive if d0 < d1, otherwise negative.
**
** e.g. between 2000-02-28 and 2001-02-28 there are 366 days
** between 2015-12-28 and 2015-12-29 there is 1 day
** between 2015-12-28 23:59:59 and 2015-12-29 00:00:01 there is 1 day
** between 2015-12-28 00:00:01 and 2015-12-28 23:59:59 there are 0 days
**
** #param {Date} d0 - start date
** #param {Date} d1 - end date
** #returns {number} - whole number of days between d0 and d1
**
*/
function daysDifference(d0, d1) {
var diff = new Date(+d1).setHours(12) - new Date(+d0).setHours(12);
return Math.round(diff/8.64e7);
}
// Simple formatter
function formatDate(date){
return [date.getFullYear(),('0'+(date.getMonth()+1)).slice(-2),('0'+date.getDate()).slice(-2)].join('-');
}
// Examples
[[new Date(2000,1,28), new Date(2001,1,28)], // Leap year
[new Date(2001,1,28), new Date(2002,1,28)], // Not leap year
[new Date(2017,0,1), new Date(2017,1,1)]
].forEach(function(dates) {
document.write('From ' + formatDate(dates[0]) + ' to ' + formatDate(dates[1]) +
' is ' + daysDifference(dates[0],dates[1]) + ' days<br>');
});
<html lang="en">
<head>
<script>
function getDateDiff(time1, time2) {
var str1= time1.split('/');
var str2= time2.split('/');
// yyyy , mm , dd
var t1 = new Date(str1[2], str1[0]-1, str1[1]);
var t2 = new Date(str2[2], str2[0]-1, str2[1]);
var diffMS = t1 - t2;
console.log(diffMS + ' ms');
var diffS = diffMS / 1000;
console.log(diffS + ' ');
var diffM = diffS / 60;
console.log(diffM + ' minutes');
var diffH = diffM / 60;
console.log(diffH + ' hours');
var diffD = diffH / 24;
console.log(diffD + ' days');
alert(diffD);
}
//alert(getDateDiff('10/18/2013','10/14/2013'));
</script>
</head>
<body>
<input type="button"
onclick="getDateDiff('10/18/2013','10/14/2013')"
value="clickHere()" />
</body>
</html>
use Moment.js for all your JavaScript related date-time calculation
Answer to your question is:
var a = moment([2007, 0, 29]);
var b = moment([2007, 0, 28]);
a.diff(b) // 86400000
Complete details can be found here
adding to #paresh mayani 's answer, to work like Facebook - showing how much time has passed in sec/min/hours/weeks/months/years
var DateDiff = {
inSeconds: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/1000);
},
inMinutes: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/60000);
},
inHours: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/3600000);
},
inDays: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000));
},
inWeeks: function(d1, d2) {
var t2 = d2.getTime();
var t1 = d1.getTime();
return parseInt((t2-t1)/(24*3600*1000*7));
},
inMonths: function(d1, d2) {
var d1Y = d1.getFullYear();
var d2Y = d2.getFullYear();
var d1M = d1.getMonth();
var d2M = d2.getMonth();
return (d2M+12*d2Y)-(d1M+12*d1Y);
},
inYears: function(d1, d2) {
return d2.getFullYear()-d1.getFullYear();
}
}
var dString = "May, 20, 1984"; //will also get (Y-m-d H:i:s)
var d1 = new Date(dString);
var d2 = new Date();
var timeLaps = DateDiff.inSeconds(d1, d2);
var dateOutput = "";
if (timeLaps<60)
{
dateOutput = timeLaps+" seconds";
}
else
{
timeLaps = DateDiff.inMinutes(d1, d2);
if (timeLaps<60)
{
dateOutput = timeLaps+" minutes";
}
else
{
timeLaps = DateDiff.inHours(d1, d2);
if (timeLaps<24)
{
dateOutput = timeLaps+" hours";
}
else
{
timeLaps = DateDiff.inDays(d1, d2);
if (timeLaps<7)
{
dateOutput = timeLaps+" days";
}
else
{
timeLaps = DateDiff.inWeeks(d1, d2);
if (timeLaps<4)
{
dateOutput = timeLaps+" weeks";
}
else
{
timeLaps = DateDiff.inMonths(d1, d2);
if (timeLaps<12)
{
dateOutput = timeLaps+" months";
}
else
{
timeLaps = DateDiff.inYears(d1, d2);
dateOutput = timeLaps+" years";
}
}
}
}
}
}
alert (dateOutput);
With momentjs it's simple:
moment("2016-04-08").fromNow();
function DateDiff(date1, date2) {
date1.setHours(0);
date1.setMinutes(0, 0, 0);
date2.setHours(0);
date2.setMinutes(0, 0, 0);
var datediff = Math.abs(date1.getTime() - date2.getTime()); // difference
return parseInt(datediff / (24 * 60 * 60 * 1000), 10); //Convert values days and return value
}
var d1=new Date(2011,0,1); // jan,1 2011
var d2=new Date(); // now
var diff=d2-d1,sign=diff<0?-1:1,milliseconds,seconds,minutes,hours,days;
diff/=sign; // or diff=Math.abs(diff);
diff=(diff-(milliseconds=diff%1000))/1000;
diff=(diff-(seconds=diff%60))/60;
diff=(diff-(minutes=diff%60))/60;
days=(diff-(hours=diff%24))/24;
console.info(sign===1?"Elapsed: ":"Remains: ",
days+" days, ",
hours+" hours, ",
minutes+" minutes, ",
seconds+" seconds, ",
milliseconds+" milliseconds.");
I think this should do it.
let today = new Date();
let form_date=new Date('2019-10-23')
let difference=form_date>today ? form_date-today : today-form_date
let diff_days=Math.floor(difference/(1000*3600*24))
based on javascript runtime prototype implementation you can use simple arithmetic to subtract dates as in bellow
var sep = new Date(2020, 07, 31, 23, 59, 59);
var today = new Date();
var diffD = Math.floor((sep - today) / (1000 * 60 * 60 * 24));
console.log('Day Diff: '+diffD);
the difference return answer as milliseconds, then you have to convert it by division:
by 1000 to convert to second
by 1000×60 convert to minute
by 1000×60×60 convert to hour
by 1000×60×60×24 convert to day
function DateDiff(b, e)
{
let
endYear = e.getFullYear(),
endMonth = e.getMonth(),
years = endYear - b.getFullYear(),
months = endMonth - b.getMonth(),
days = e.getDate() - b.getDate();
if (months < 0)
{
years--;
months += 12;
}
if (days < 0)
{
months--;
days += new Date(endYear, endMonth, 0).getDate();
}
return [years, months, days];
}
[years, months, days] = DateDiff(
new Date("October 21, 1980"),
new Date("July 11, 2017")); // 36 8 20
Sorry but flat millisecond calculation is not reliable
Thanks for all the responses, but few of the functions I tried are failing either on
1. A date near today's date
2. A date in 1970 or
3. A date in a leap year.
Approach that best worked for me and covers all scenario e.g. leap year, near date in 1970, feb 29 etc.
var someday = new Date("8/1/1985");
var today = new Date();
var years = today.getFullYear() - someday.getFullYear();
// Reset someday to the current year.
someday.setFullYear(today.getFullYear());
// Depending on when that day falls for this year, subtract 1.
if (today < someday)
{
years--;
}
document.write("Its been " + years + " full years.");
This code will return the difference between two dates in days:
const previous_date = new Date("2019-12-23");
const current_date = new Date();
const current_year = current_date.getFullYear();
const previous_date_year =
previous_date.getFullYear();
const difference_in_years = current_year -
previous_date_year;
let months = current_date.getMonth();
months = months + 1; // for making the indexing
// of months from 1
for(let i = 0; i < difference_in_years; i++){
months = months + 12;
}
let days = current_date.getDate();
days = days + (months * 30.417);
console.log(`The days between ${current_date} and
${previous_date} are : ${days} (approximately)`);
If you are using moment.js then it is pretty simple to find date difference.
var now = "04/09/2013 15:00:00";
var then = "04/09/2013 14:20:30";
moment.utc(moment(now,"DD/MM/YYYY HH:mm:ss").diff(moment(then,"DD/MM/YYYY HH:mm:ss"))).format("HH:mm:ss")
This is how you can implement difference between dates without a framework.
function getDateDiff(dateOne, dateTwo) {
if(dateOne.charAt(2)=='-' & dateTwo.charAt(2)=='-'){
dateOne = new Date(formatDate(dateOne));
dateTwo = new Date(formatDate(dateTwo));
}
else{
dateOne = new Date(dateOne);
dateTwo = new Date(dateTwo);
}
let timeDiff = Math.abs(dateOne.getTime() - dateTwo.getTime());
let diffDays = Math.ceil(timeDiff / (1000 * 3600 * 24));
let diffMonths = Math.ceil(diffDays/31);
let diffYears = Math.ceil(diffMonths/12);
let message = "Difference in Days: " + diffDays + " " +
"Difference in Months: " + diffMonths+ " " +
"Difference in Years: " + diffYears;
return message;
}
function formatDate(date) {
return date.split('-').reverse().join('-');
}
console.log(getDateDiff("23-04-2017", "23-04-2018"));
function daysInMonth (month, year) {
return new Date(year, month, 0).getDate();
}
function getduration(){
let A= document.getElementById("date1_id").value
let B= document.getElementById("date2_id").value
let C=Number(A.substring(3,5))
let D=Number(B.substring(3,5))
let dif=D-C
let arr=[];
let sum=0;
for (let i=0;i<dif+1;i++){
sum+=Number(daysInMonth(i+C,2019))
}
let sum_alter=0;
for (let i=0;i<dif;i++){
sum_alter+=Number(daysInMonth(i+C,2019))
}
let no_of_month=(Number(B.substring(3,5)) - Number(A.substring(3,5)))
let days=[];
if ((Number(B.substring(3,5)) - Number(A.substring(3,5)))>0||Number(B.substring(0,2)) - Number(A.substring(0,2))<0){
days=Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter
}
if ((Number(B.substring(3,5)) == Number(A.substring(3,5)))){
console.log(Number(B.substring(0,2)) - Number(A.substring(0,2)) + sum_alter)
}
time_1=[]; time_2=[]; let hour=[];
time_1=document.getElementById("time1_id").value
time_2=document.getElementById("time2_id").value
if (time_1.substring(0,2)=="12"){
time_1="00:00:00 PM"
}
if (time_1.substring(9,11)==time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))
}
if (time_1.substring(9,11)!=time_2.substring(9,11)){
hour=Math.abs(Number(time_2.substring(0,2)) - Number(time_1.substring(0,2)))+12
}
let min=Math.abs(Number(time_1.substring(3,5))-Number(time_2.substring(3,5)))
document.getElementById("duration_id").value=days +" days "+ hour+" hour " + min+" min "
}
<input type="text" id="date1_id" placeholder="28/05/2019">
<input type="text" id="date2_id" placeholder="29/06/2019">
<br><br>
<input type="text" id="time1_id" placeholder="08:01:00 AM">
<input type="text" id="time2_id" placeholder="00:00:00 PM">
<br><br>
<button class="text" onClick="getduration()">Submit </button>
<br><br>
<input type="text" id="duration_id" placeholder="days hour min">
var date1 = new Date("06/30/2019");
var date2 = new Date("07/30/2019");
// To calculate the time difference of two dates
var Difference_In_Time = date2.getTime() - date1.getTime();
// To calculate the no. of days between two dates
var Difference_In_Days = Difference_In_Time / (1000 * 3600 * 24);
//To display the final no. of days (result)
document.write("Total number of days between dates <br>"
+ date1 + "<br> and <br>"
+ date2 + " is: <br> "
+ Difference_In_Days);
this should work just fine if you just need to show what time left, since JavaScript uses frames for its time you'll have get your End Time - The Time RN after that we can divide it by 1000 since apparently 1000 frames = 1 seconds, after that you can use the basic math of time, but there's still a problem to this code, since the calculation is static, it can't compensate for the different day total in a year (360/365/366), the bunch of IF after the calculation is to make it null if the time is lower than 0, hope this helps even though it's not exactly what you're asking :)
var now = new Date();
var end = new Date("End Time");
var total = (end - now) ;
var totalD = Math.abs(Math.floor(total/1000));
var years = Math.floor(totalD / (365*60*60*24));
var months = Math.floor((totalD - years*365*60*60*24) / (30*60*60*24));
var days = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24)/ (60*60*24));
var hours = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24)/ (60*60));
var minutes = Math.floor((totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60)/ (60));
var seconds = Math.floor(totalD - years*365*60*60*24 - months*30*60*60*24 - days*60*60*24 - hours*60*60 - minutes*60);
var Y = years < 1 ? "" : years + " Years ";
var M = months < 1 ? "" : months + " Months ";
var D = days < 1 ? "" : days + " Days ";
var H = hours < 1 ? "" : hours + " Hours ";
var I = minutes < 1 ? "" : minutes + " Minutes ";
var S = seconds < 1 ? "" : seconds + " Seconds ";
var A = years == 0 && months == 0 && days == 0 && hours == 0 && minutes == 0 && seconds == 0 ? "Sending" : " Remaining";
document.getElementById('txt').innerHTML = Y + M + D + H + I + S + A;
Ok, there are a bunch of ways you can do that.
Yes, you can use plain old JS. Just try:
let dt1 = new Date()
let dt2 = new Date()
Let's emulate passage using Date.prototype.setMinutes and make sure we are in range.
dt1.setMinutes(7)
dt2.setMinutes(42)
console.log('Elapsed seconds:',(dt2-dt1)/1000)
Alternatively you could use some library like js-joda, where you can easily do things like this (directly from docs):
var dt1 = LocalDateTime.parse("2016-02-26T23:55:42.123");
var dt2 = dt1
.plusYears(6)
.plusMonths(12)
.plusHours(2)
.plusMinutes(42)
.plusSeconds(12);
// obtain the duration between the two dates
dt1.until(dt2, ChronoUnit.YEARS); // 7
dt1.until(dt2, ChronoUnit.MONTHS); // 84
dt1.until(dt2, ChronoUnit.WEEKS); // 356
dt1.until(dt2, ChronoUnit.DAYS); // 2557
dt1.until(dt2, ChronoUnit.HOURS); // 61370
dt1.until(dt2, ChronoUnit.MINUTES); // 3682242
dt1.until(dt2, ChronoUnit.SECONDS); // 220934532
There are plenty more libraries ofc, but js-joda has an added bonus of being available also in Java, where it has been extensively tested. All those tests have been migrated to js-joda, it's also immutable.
I made a below function to get the difference between now and "2021-02-26T21:50:42.123".
The difference return answer as milliseconds, so I convert it by using this formula:
(1000 * 3600 * 24).
function getDiff(dateAcquired) {
let calDiff = Math.floor(
(new Date() - new Date(dateAcquired)) / (1000 * 3600 * 24)
);
return calDiff;
}
console.log(getDiff("2021-02-26T21:50:42.123"));
Can be useful :
const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/24 * 60 * 60 * 1000)
or
const date_diff = (date1, date2) => Math.ceil(Math.abs(date1 - date2)/86400000)
where 24 * 60 * 60 * 1000 is (day * minutes * seconds * milliseconds) = 86400000 milliseconds in one day
Thank you
// the idea is to get time left for new year.
// Not considering milliseconds as of now, but that
// can be done
var newYear = '1 Jan 2023';
const secondsInAMin = 60;
const secondsInAnHour = 60 * secondsInAMin;
const secondsInADay = 24 * secondsInAnHour;
function DateDiffJs() {
var newYearDate = new Date(newYear);
var currDate = new Date();
var remainingSecondsInDateDiff = (newYearDate - currDate) / 1000;
var days = Math.floor(remainingSecondsInDateDiff / secondsInADay);
var remainingSecondsAfterDays = remainingSecondsInDateDiff - (days * secondsInADay);
var hours = Math.floor(remainingSecondsAfterDays / secondsInAnHour);
var remainingSecondsAfterhours = remainingSecondsAfterDays - (hours * secondsInAnHour);
var mins = Math.floor(remainingSecondsAfterhours / secondsInAMin);
var seconds = Math.floor(remainingSecondsAfterhours - (mins * secondsInAMin));
console.log(`days :: ${days}`)
console.log(`hours :: ${hours}`)
console.log(`mins :: ${mins}`)
console.log(`seconds :: ${seconds}`)
}
DateDiffJs();
I tried the JS below:
var start = new Date("25-05-2016");
var finish = new Date("31-05-2016");
var dayMilliseconds = 1000 * 60 * 60 * 24;
var weekendDays = 0;
while (start <= finish) {
var day = start.getDay()
if (day == 0) {
weekendDays++;
}
start = new Date(+start + dayMilliseconds);
}
alert(weekendDays);
However, it gives the wrong output.
I need to get the total count of Sundays between the two dates.
You use incorrect date format.It will work if init date so:
var start = new Date("2016-05-25");
var finish = new Date("2016-05-31");
Your date format is wrong. Dates' string format is "yyyy-mm-dd". See here for more information.
Also, looping each day of the interval is very inefficient. You may try the following instead.
function getNumberOfWeekDays(start, end, dayNum){
// Sunday's num is 0 with Date.prototype.getDay.
dayNum = dayNum || 0;
// Calculate the number of days between start and end.
var daysInInterval = Math.ceil((end.getTime() - start.getTime()) / (1000 * 3600 * 24));
// Calculate the nb of days before the next target day (e.g. next Sunday after start).
var toNextTargetDay = (7 + dayNum - start.getDay()) % 7;
// Calculate the number of days from the first target day to the end.
var daysFromFirstTargetDay = Math.max(daysInInterval - toNextTargetDay, 0);
// Calculate the number of weeks (even partial) from the first target day to the end.
return Math.ceil(daysFromFirstTargetDay / 7);
}
var start = new Date("2016-05-25");
var finish = new Date("2016-05-31");
console.log("Start:", start);
console.log("Start's week day num:", start.getDay());
console.log("Finish:", finish);
console.log("Finish's week day num:", finish.getDay());
console.log("Number of Sundays:", getNumberOfWeekDays(start, finish));
Your date format and comparison condition should change like the following:
var start = new Date("2016-05-11");
var finish = new Date("2016-05-31");
var dayMilliseconds = 1000 * 60 * 60 * 24;
var weekendDays = 0;
while (start.getTime() <= finish.getTime()) {
var day = start.getDay();
if (day == 0) {
weekendDays++;
}
start = new Date(+start + dayMilliseconds);
}
alert(weekendDays);
Check Fiddle
You are using incorrect date format.
Just Change the format to:
var start = new Date(2016, 4, 25);
var finish = new Date(2016, 4, 31);
Try this function:
function CalculateWeekendDays(fromDate, toDate){
var weekendDayCount = 0;
while(fromDate < toDate){
fromDate.setDate(fromDate.getDate() + 1);
if(fromDate.getDay() === 0){
++weekendDayCount ;
}
}
return weekendDayCount ;
}
console.log(CalculateWeekendDays(new Date(2011, 6, 2), new Date(2011, 7, 2)));
This will give you number of sunday come between 2 dates
change your date format.It will work
var start = new Date("05-16-2016");
var finish = new Date("05-31-2016");
var dayMilliseconds = 1000 * 60 * 60 * 24;
var weekendDays = 0;
while (start <= finish) {
var day = start.getDay()
if (day == 0) {
weekendDays++;
}
start = new Date(+start + dayMilliseconds);
}
console.log(weekendDays);
JS date format doesn't have "dd-MM-yyyy" ,so it will invalid date format .Try recreate date is ok or just change your date format Date Format
Try this:
var start = new Date("25-05-2016");
var end = new Date("31-05-2016");
var startDate = new Date(start);
var endDate = new Date(end);
var totalSundays = 0;
for (var i = startDate; i <= endDate; ){
if (i.getDay() == 0){
totalSundays++;
}
i.setTime(i.getTime() + 1000*60*60*24);
}
console.log(totalSundays);
// Find date of sundays b/w two dates
var fromDate = new Date('2022-10-26')
var toDate = new Date('2022-11-31')
var sunday = 0
var milisec = 1000 * 60 * 60 * 24;
while (fromDate <= toDate) {
var day = fromDate.getDay()
if (day == 0) {
sunday++
console.log('Date of sunday:', fromDate)
}
fromDate = new Date(+fromDate + milisec)
}
console.log('Total no. of sundays:', sunday)
Is there a way to check if a date is less than 1 hour ago like this?
// old date
var olddate = new Date("February 9, 2012, 12:15");
// current date
var currentdate = new Date();
if (olddate >= currentdate - 1 hour) {
alert("newer than 1 hour");
else {
alert("older than 1 hour");
}
Also, different question - is there a way to add hours to a date like this?
var olddate = new Date("February 9, 2012, 12:15") + 15 HOURS; // output: February 10, 2012, 3:15
Define
var ONE_HOUR = 60 * 60 * 1000; /* ms */
then you can do
((new Date) - myDate) < ONE_HOUR
To get one hour from a date, try
new Date(myDate.getTime() + ONE_HOUR)
Using some ES6 syntax:
const lessThanOneHourAgo = (date) => {
const HOUR = 1000 * 60 * 60;
const anHourAgo = Date.now() - HOUR;
return date > anHourAgo;
}
Using the Moment library:
const lessThanOneHourAgo = (date) => {
return moment(date).isAfter(moment().subtract(1, 'hours'));
}
Shorthand syntax with Moment:
const lessThanOneHourAgo = (date) => moment(date).isAfter(moment().subtract(1, 'hours'));
the moment library can really help express this. The trick is to take the date, add time, and see if it's before or after now:
lastSeenAgoLabel: function() {
var d = this.lastLogin();
if (! moment(d).isValid()) return 'danger'; // danger if not a date.
if (moment(d).add(10, 'minutes').isBefore(/*now*/)) return 'danger'; // danger if older than 10 mins
if (moment(d).add(5, 'minutes').isBefore(/*now*/)) return 'warning'; // warning if older than 5mins
return 'success'; // Looks good!
},
Using moment will be much easier in this case, You could try this:
let hours = moment().diff(moment(yourDateString), 'hours');
It will give you integer value like 1,2,5,0etc so you can easily use condition check like:
if(hours < 1) {
Also, one more thing is you can get more accurate result of the time difference (in decimals like 1.2,1.5,0.7etc) to get this kind of result use this syntax:
let hours = moment().diff(moment(yourDateString), 'hours', true);
Let me know if you have any further query
//for adding hours to a date
Date.prototype.addHours= function(hrs){
this.setHours(this.getHours()+hrs);
return this;
}
Call function like this:
//test alert(new Date().addHours(4));
You can do it as follows:
First find difference of two dates i-e in milliseconds
Convert milliseconds into minutes
If minutes are less than 60, then it means date is within hour else not within hour.
var date = new Date("2020-07-12 11:30:10");
var now = new Date();
var diffInMS = now - date;
var msInHour = Math.floor(diffInMS/1000/60);
if (msInHour < 60) {
console.log('Within hour');
} else {
console.log('Not within the hour');
}
Plain JavaScript solution with in 12 days and 12 days ago option
const timeAgo = ( inputDate ) => {
const date = ( inputDate instanceof Date) ? inputDate : new Date(inputDate);
const FORMATTER = new Intl.RelativeTimeFormat('en');
const RANGES = {
years : 3600 * 24 * 365,
months : 3600 * 24 * 30,
weeks : 3600 * 24 * 7,
days : 3600 * 24,
hours : 3600,
minutes : 60,
seconds : 1
};
const secondsElapsed = (date.getTime() - Date.now()) / 1000;
for (let key in RANGES) {
if ( RANGES[key] < Math.abs(secondsElapsed) ) {
const delta = secondsElapsed / RANGES[key];
return FORMATTER.format(Math.round(delta), key);
}
}
}
// OUTPUTS
console.log( timeAgo('2040-12-24') )
console.log( timeAgo('6 Sept, 2012') );
console.log( timeAgo('2022-05-27T17:45:01+0000') );
let d = new Date()
console.log( "Date will change: ", timeAgo( d.setHours(24,0,0,0) ) );
// d.setDate( d.getDate() - 0 );
d.setHours(-24,0,0,0); // (H,M,S,MS) | 24 hours format
console.log("Day started: " , timeAgo( d ) );
//try this:
// to compare two date's:
<Script Language=Javascript>
function CompareDates()
{
var str1 = document.getElementById("Fromdate").value;
var str2 = document.getElementById("Todate").value;
var dt1 = parseInt(str1.substring(0,2),10);
var mon1 = parseInt(str1.substring(3,5),10);
var yr1 = parseInt(str1.substring(6,10),10);
var dt2 = parseInt(str2.substring(0,2),10);
var mon2 = parseInt(str2.substring(3,5),10);
var yr2 = parseInt(str2.substring(6,10),10);
var date1 = new Date(yr1, mon1, dt1);
var date2 = new Date(yr2, mon2, dt2);
if(date2 < date1)
{
alert("To date cannot be greater than from date");
return false;
}
else
{
alert("Submitting ...");
}
}
</Script>
Hope it will work 4 u...
I'm passing my calendar selected date of birth to following JS function for calculating Age:
var DOBmdy = date.split("-");
Bdate = new Date(DOBmdy[2],DOBmdy[0]-1,DOBmdy[1]);
BDateArr = (''+Bdate).split(' ');
//document.getElementById('DOW').value = BDateArr[0];
Cdate = new Date;
CDateArr = (''+Cdate).split(" ");
Age = CDateArr[3] - BDateArr[3];
Now, lets say, input age is: 2nd Aug 1983 and age count comes: 28, while as August month has not been passed yet, i want to show the current age of 27 and not 28
Any idea, how can i write that logic, to count age 27 perfectly with my JS function.
Thanks !
Let birth date be august 2nd 1983, then the difference in milliseconds between now an that date is:
var diff = new Date - new Date('1983-08-02');
The difference in days is (1 second = 1000 ms, 1 hour = 60*60 seconds, 1 day = 24 * 1 hour)
var diffdays = diff / 1000 / (60 * 60 * 24);
The difference in years (so, the age) becomes (.25 to account for leapyears):
var age = Math.floor(diffdays / 365.25);
Now try it with
diff = new Date('2011-08-01') - new Date('1983-08-02'); //=> 27
diff = new Date('2011-08-02') - new Date('1983-08-02'); //=> 28
diff = new Date('2012-08-02') - new Date('1983-08-02'); //=> 29
So, your javascript could be rewritten as:
var Bdate = new Date(date.split("-").reverse().join('-')),
age = Math.floor( ( (Cdate - Bdate) / 1000 / (60 * 60 * 24) ) / 365.25 );
[edit] Didn't pay enough attention. date.split('-') gives the array [dd,mm,yyyy], so reversing it results in[yyyy,mm,dd]. Now joining that again using '-', the result is the string 'yyyy-mm-dd', which is valid input for a new Date.
(new Date() - new Date('08-02-1983')) / 1000 / 60 / 60 / 24 / 365.25
That will get you the difference in years, you will occasionally run into off-by-one-day issues using this.
May be this works:
var today = new Date();
var d = document.getElementById("dob").value;
if (!/\d{4}\-\d{2}\-\d{2}/.test(d)) { // check valid format
return false;
}
d = d.split("-");
var byr = parseInt(d[0]);
var nowyear = today.getFullYear();
if (byr >= nowyear || byr < 1900) { // check valid year
return false;
}
var bmth = parseInt(d[1],10)-1;
if (bmth<0 || bmth>11) { // check valid month 0-11
return false;
}
var bdy = parseInt(d[2],10);
if (bdy<1 || bdy>31) { // check valid date according to month
return false;
}
var age = nowyear - byr;
var nowmonth = today.getMonth();
var nowday = today.getDate();
if (bmth > nowmonth) {age = age - 1} // next birthday not yet reached
else if (bmth == nowmonth && nowday < bdy) {age = age - 1}
alert('You are ' + age + ' years old');
I just had to write a function to do this and thought'd I'd share.
This is accurate from a human point of view! None of that crazy 365.2425 stuff.
var ageCheck = function(yy, mm, dd) {
// validate input
yy = parseInt(yy,10);
mm = parseInt(mm,10);
dd = parseInt(dd,10);
if(isNaN(dd) || isNaN(mm) || isNaN(yy)) { return 0; }
if((dd < 1 || dd > 31) || (mm < 1 || mm > 12)) { return 0; }
// change human inputted month to javascript equivalent
mm = mm - 1;
// get today's date
var today = new Date();
var t_dd = today.getDate();
var t_mm = today.getMonth();
var t_yy = today.getFullYear();
// We are using last two digits, so make a guess of the century
if(yy == 0) { yy = "00"; }
else if(yy < 9) { yy = "0"+yy; }
yy = (today.getFullYear() < "20"+yy ? "19"+yy : "20"+yy);
// Work out the age!
var age = t_yy - yy - 1; // Starting point
if( mm < t_mm ) { age++;} // If it's past their birth month
if( mm == t_mm && dd <= t_dd) { age++; } // If it's past their birth day
return age;
}